In every power of 3 the tens digit is an even number

Use induction on the exponent:

The last digit is either 1, 3, 7 or 9.

If 1 or 3, multiplying a power of 3 will not change the evenness of the second last digit.

If 7 or 9, multiplying will result in a carry over of 2, which again does not change the evenness of the second last digit.

Hint $\,\bmod 20\!:\,\ 3^{\large 4} \equiv 1\,\Rightarrow\,3^{\large n}\equiv 3^{\,\large n\bmod 4}\in \{\color{#c00}01,\color{#c00}03,\color{#c00}09,\color{#c00}27\}$

Inductively: $\ 3^{\large n} = 20\,k\, +\{0,3,9,7\}\,\ $

$\qquad\ \Rightarrow\ 3^{\large n+1} = 20 K+\{0,9,7,1\},\,\ K = 3k\!+\!i,\,\ i\in \{ 0,1\}$

Notice $\,r< 10\,\Rightarrow\, 20k+r = (\color{#0a0}{2k})10 + r\,$ has even tens digits $\,\color{#0a0}{2k} \bmod 10 = 2(k\bmod 5)$

Moron's answer is very elegant.

Here is an easy but tedious way of dealing with many similar questions: For any $n$ and $k$, the last $k$ digits of the numbers in the sequence $n^1,n^2,n^3,\dots$ eventually repeat, so we only need to check a few cases.

For the question at hand, note that $3^2=9=10-1$, so $3^{20}=(10-1)^{10}$ has the form $100k+1$ for some $k$. This means that the last 2 digits of powers of 3 repeat each 20 numbers. So we just need to check the last two digits of $3^0,3^1,\dots,3^{19}$, and this can be done very quickly even by hand: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$

Case 1: $n = 2k; k$ even.

$3^{2k} = 9^{k} = (10 - 1)^{k} = 10^{k} - {k \choose 2}^{k-1} + .... + {k \choose k-2}10^2 - {k \choose k - 1}10 + 1 = N*100 - k*10 + 1$. $k$ is even so the second digit is even.

Case 2: $n = 2k; k$ odd.

$3^{2k} = 9^{k} = (10 - 1)^{k} = 10^{k} - {k \choose 2}^{k-1} + .... - {k \choose k-2}10^2 + {k \choose k - 1}10 - 1 = N*100 + k*10 - 1$. $k$ is odd so $k*10 - 1 = (k - 1)*10 + 9$ so the second digit is even.

Case 3: $n = 2k + 1; k$ even.

$3^{2k} = 9^{k} = (10 - 1)^{k} = 10^{k} - {k \choose 2}^{k-1} + .... + {k \choose k-2}10^2 - {k \choose k - 1}10 + 1 = N*100 - k*10 + 1$.

$3^{2k+1} = $M*100 - 3k*10 + 3$. As $k$ is even the second digit is even.

Case 4: $n = 2k + 1; k$ odd.

$3^{2k} = 9^{k} = (10 - 1)^{k} = 10^{k} - {k \choose 2}^{k-1} + .... - {k \choose k-2}10^2 + {k \choose k - 1}10 - 1 = N*100 + k*10 - 1$.

$3^{2k + 1} = M*100 + 3k*10 - 7 = M*100 + (3k -1)*10 + 7$. As $3k -1$ is even, the second digit is even.

This also tells you what the first digit is: 1, 9, 3, or 7. Which gives a hint that might be a more efficient method.

Method 2:

$3^4 = 81 = 8*10 + 1$

So if $n = 4m + k; k = 0,1,2,3$

$3^n = (3^4)^m 3^k = (8*10 + 1)^m 3^k = (M*100 + 8*m*10 + 1)\{1,3,9,27\}$

$= N*100 + 8*m*\{1,3,9,7\}*10 \{+ 2*10\} + 1*\{1,3,9,7\}$

Or more elegantly as Bill Dubuque pointed out

$3^4 = 1 \mod (20) \implies 3^{4m + k} = (3^4)^m 3^k = 3^k \mod 20 \implies$

$ 3^n = \{1,3,9,7\} \mod 20$.